Digital Search Trees & Binary Tries Analog of radix sort to searching. Keys are binary bit strings. Fixed length – 0110, 0010, 1010, 1011. Variable length – 01, 00, 101, 1011. Application – IP routing, packet classification, firewalls. IPv4 – 32 bit IP address; |key| <= 32. IPv6 – 128 bit IP address; |key| <= 128. Will discuss IP application in detail later. IPv4 (v6) the keys in the search structure are up to 32 bits (128) long. Application of variable length binary keysHuffman codes/tree.

Digital Search Tree Assume fixed number of bits. Not empty => Root contains one dictionary pair (any pair). All remaining pairs whose key begins with a 0 are in the left subtree. All remaining pairs whose key begins with a 1 are in the right subtree. Left and right subtrees are digital search trees on remaining bits. So, on level I branching is done on bit I of the key.

Example Start with an empty digital search tree and insert a pair whose key is 0110. 0110 Now, insert a pair whose key is 0010. 0110 0010

Example Now, insert a pair whose key is 1001. 0110 0010 1001 0110 0010

Example Now, insert a pair whose key is 1011. 1001 0110 0010 1011 1001

Example Now, insert a pair whose key is 0000. 1001 0110 0010 1011 0000

Search/Insert/Delete 1001 0110 0010 1011 0000 Delete from nonleaf => replace with key in subtree leaf. Note can’t delete as in binary search tree as that would change levels of nodes and mess up relationship between level and branching bit. Complexity of each operation is O(#bits in a key). #key comparisons = O(height). Expensive when keys are very long.

Binary Trie Information Retrieval. At most one key comparison per operation. Fixed length keys. Branch nodes. Left and right child pointers. No data field(s). Element nodes. No child pointers. Data field to hold dictionary pair. Huffman trees have same structure but for variable length keys that satisfy the prefix property (no key is a proper prefix of another)

Example At most one key comparison for a search. 1 1100 0001 0011 1000 1001 1 Every branch node has >=2 keys/pairs in its subtree. At most one key comparison for a search.

Variable Key Length Left and right child fields. Left and right pair fields. Left pair is pair whose key terminates at root of left subtree or the single pair that might otherwise be in the left subtree. Right pair is pair whose key terminates at root of right subtree or the single pair that might otherwise be in the right subtree. Field is null otherwise.

Example At most one key comparison for a search. null 1 00 01100 10 null 1 00 01100 10 11111 0000 001 1000 101 1 Alternative is to store only length k keys at level k. Then, 0000, for example, would be stored one level down from shown figure. Longest prefix matching. 00100 001100 At most one key comparison for a search.

Fixed Length Insert Insert 0111. Zero compares. 1 1 1100 0111 0001 0011 1100 1000 1001 1 0111 1 Insert 0111. Zero compares.

Fixed Length Insert 0001 0011 1100 1000 1001 1 0111 1 Insert 1101.

Fixed Length Insert 1100 0001 0011 1000 1001 1 0111 1 Insert 1101.

Fixed Length Insert Insert 1101. One compare. 0001 0011 1000 1001 1 1 0111 1 1 1100 1101 Insert 1101. One compare.

Fixed Length Delete Delete 0111. 1 0001 0011 1000 1001 0111 1100 1101 1 0001 0011 1000 1001 0111 1100 1101 Sibling is branch node. Delete 0111.

Fixed Length Delete Delete 0111. One compare. 1 0001 0011 1000 1001 1 0001 0011 1000 1001 1100 1101 Delete 0111. One compare.

Fixed Length Delete Delete 1100. 1 0001 0011 1000 1001 1100 1101 1 0001 0011 1000 1001 1100 1101 Sibling is element node. Delete 1100.

Fixed Length Delete 1 1 1 0001 0011 1 1 1000 1001 1101 Delete 1100.

Fixed Length Delete 1101 1 1 1 0001 0011 1 1000 1001 Delete 1100.

Fixed Length Delete 1101 1 1 1 0001 0011 1 1000 1001 Delete 1100.

Fixed Length Delete Delete 1100. One compare. 1 1 1101 1 0001 0011 1 1 1 1101 1 0001 0011 1 1000 1001 Delete 1100. One compare.

Fixed Length Join(S,m,B) Convert 3-way join to 2-way join Insert m into B to get B’. S empty => B’ is answer; done. S is element node => insert S element into B’; done; B’ is element node => insert B’ element into S; done; If you get to this step, the roots of S and B’ are branch nodes.

Fixed Length Join(S,m,B) S has empty right subtree. a S b c B’ c J(S,B’) J(a,b) + = J(X,Y) ajoin X and Y, all keys in X < all in Y.

Fixed Length Join(S,m,B) S has nonempty right subtree. Left subtree of B’ must be empty, because all keys in B’ > all keys in S. a S b c B’ a J(S,B’) J(b,c) + = Complexity = O(height).